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3 years ago

Which statement below is not a premise of the Kinetic-molecular theory of gases?

Chemistry

1 answer:

zhuklara [117]3 years ago

4 0

Answer:

d = all are premise of KMT.

Explanation:

Gas particles are extremely small and have relatively large distance between them.

Gas particles are continuously moving in random, straight-line motion as they are collide with each other and the container walls.

The average kinetic energy of gas particles is proportional to the temperature of gas.

All these are the premise of kinetic molecular theory. According to kinetic molecular theory, the particles pf gases are very small and randomly move in the available space. They move freely in straight line and colloid with each other and also the wall of container. This collision is elastic. As molecules strike with each other and walls of container their energy increases, and they gets warmer. So their kinetic energy is proportional to the temperature because mass of particle is constant.

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3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

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